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Write down the Lagrangian for a projectile (subject to no air resistance) in terms of its Cartesian coordinates \((x, y, z),\) with \(z\) measured vertically upward. Find the three Lagrange equations and show that they are exactly what you would expect for the equations of motion.

Short Answer

Expert verified
The Lagrangian is \( L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - mgz \). The equations: \( \ddot{x} = 0 \), \( \ddot{y} = 0 \), \( \ddot{z} = -g \).

Step by step solution

01

Understanding the Lagrangian Function

The Lagrangian function, defined as \( L = T - V \), where \( T \) is the kinetic energy and \( V \) is the potential energy, is used to derive the equations of motion. For a projectile in three-dimensional space, we'll express both the kinetic and potential energy in terms of the Cartesian coordinates \((x, y, z)\).
02

Express Kinetic Energy in Cartesian Coordinates

The kinetic energy \( T \) of a projectile with mass \( m \) moving with velocity components \( \dot{x}, \dot{y}, \dot{z} \) is given by \( T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \). These are the squared velocity components in the Cartesian system.
03

Express Potential Energy in Cartesian Coordinates

The potential energy \( V \) of a projectile in a gravitational field (with constant gravity \( g \)) depends on the vertical position \( z \). It is expressed as \( V = mgz \), where \( z \) is the height above the reference point.
04

Write the Lagrangian

Substitute the expressions for kinetic and potential energy into the Lagrangian: \[ L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - mgz \] This function represents the Lagrangian of the system.
05

Derive the Lagrange Equations

Using the Euler-Lagrange equation, \( \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0 \), apply it to each coordinate \( x, y, \) and \( z \).
06

Equation for the x-direction

Applying the Lagrange equation to \( x \):\[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{x}} \right) = m\ddot{x} \quad \text{and} \quad \frac{\partial L}{\partial x} = 0 \]Thus, the equation becomes: \[ m\ddot{x} = 0 \] indicating no acceleration in the \( x \)-direction.
07

Equation for the y-direction

Similarly, apply the Lagrange equation to \( y \):\[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{y}} \right) = m\ddot{y} \quad \text{and} \quad \frac{\partial L}{\partial y} = 0 \]Thus, \[ m\ddot{y} = 0 \] indicating no acceleration in the \( y \)-direction as well.
08

Equation for the z-direction

For the \( z \) coordinate, we have:\[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{z}} \right) = m\ddot{z} \quad \text{and} \quad \frac{\partial L}{\partial z} = -mg \]This results in:\[ m\ddot{z} = -mg \] which represents the acceleration due to gravity.
09

Summary of Equations of Motion

The Lagrange equations lead to the expected equations of motion: \[ \ddot{x} = 0, \quad \ddot{y} = 0, \quad \ddot{z} = -g \] These equations show that the projectile moves with constant velocity in the \( x \) and \( y \) directions, and with constant acceleration \( -g \) in the \( z \) direction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Projectile Motion
Projectile motion describes the path of an object thrown into space, subject to gravitational forces alone. In our exercise, we're analyzing this motion without air resistance, focusing on three-dimensional space with coordinates
  • \( (x, y, z) \)
Here, \( z \) is vertical. The key factors in projectile motion are:
  • Initial Velocity: Determines how the object starts its trajectory.
  • Gravity: Acts downward, affecting vertical motion.
In our case, the Lagrange equations reveal that horizontal movements (\( x \) and \( y \)) have no acceleration. The vertical motion (\( z \)) accelerates down due to gravity (-g). This analysis gives us an understanding of how an object like a ball thrown from a cliff behaves under gravity's influence.
Euler-Lagrange Equation
The Euler-Lagrange equation is a fundamental principle to derive equations of motion in Lagrangian mechanics. It connects the Lagrangian function to physical trajectories of systems. The equation is:\[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0\]Where \( L = T - V \), it represents the difference between kinetic and potential energy. Here’s how it applies:
  • For each coordinate (\( x, y, z \)): We take partial derivatives related to velocity and position.
  • Motion Derivation: This leads to equations describing how the system evolves over time.
In our projectile exercise, applying this to \( x \) and \( y \) showed no force acting horizontally thus no acceleration (\( \ddot{x} = 0 \)). But for \( z \), gravity force exists, leading to \( \ddot{z} = -g \), reflecting downward acceleration. Understanding the Euler-Lagrange equation lets us derive these natural laws from simple principles.
Kinetic and Potential Energy
Kinetic and potential energies are key to understanding Lagrangian mechanics. They are components of the Lagrangian function \( L = T - V \), representing how energy changes in motion.

Kinetic Energy (T)

Kinetic energy is associated with the motion of the projectile. Given by:\[ T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \]
  • Mass \( m \): How heavy the object is.
  • Velocity Components (\( \dot{x}, \dot{y}, \dot{z} \)): Speed in the three axes.
These factors combine to show how fast an object is moving in the Cartesian plane.

Potential Energy (V)

Potential energy relates to the object's position in a gravitational field:\[ V = mgz\]
  • Gravitational Force (\( g \)): Constant pulling the object down.
  • Height (\( z \)): Position elevated above a reference point.
Thus, an object's potential energy changes as it moves vertically. Understanding these energies helps to reflect the trade-offs in motion, vital for predicting the projectile's path.

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Most popular questions from this chapter

A mass \(m_{1}\) rests on a frictionless horizontal table and is attached to a massless string. The string runs horizontally to the edge of the table, where it passes over a massless, frictionless pulley and then hangs vertically down. A second mass \(m_{2}\) is now attached to the bottom end of the string. Write down the Lagrangian for the system. Find the Lagrange equation of motion, and solve it for the acceleration of the blocks. For your generalized coordinate, use the distance \(x\) of the second mass below the tabletop.

A pendulum is made from a massless spring (force constant \(k\) and unstretched length \(l_{\mathrm{o}}\)) that is suspended at one end from a fixed pivot \(O\) and has a mass \(m\) attached to its other end. The spring can stretch and compress but cannot bend, and the whole system is confined to a single vertical plane. (a) Write down the Lagrangian for the pendulum, using as generalized coordinates the usual angle \(\phi\) and the length \(r\) of the spring. (b) Find the two Lagrange equations of the system and interpret them in terms of Newton's second law, as given in Equation (1.48). (c) The equations of part (b) cannot be solved analytically in general. However, they can be solved for small oscillations. Do this and describe the motion. [Hint: Let \(l\) denote the equilibrium length of the spring with the mass hanging from it and write \(r=l+\epsilon .\) "Small oscillations" involve only small values of \(\epsilon\) and \(\phi,\) so you can use the small-angle approximations and drop from your equations all terms that involve powers of \(\epsilon\) or \(\phi\) (or their derivatives) higher than the first power (also products of \(\epsilon\) and \(\phi\) or their derivatives). This dramatically simplifies and uncouples the equations.]

A small cart (mass \(m\) ) is mounted on rails inside a large cart. The two are attached by a spring (force constant \(k\) ) in such a way that the small cart is in equilibrium at the midpoint of the large. The distance of the small cart from its equilibrium is denoted \(x\) and that of the large one from a fixed point on the ground is \(X,\) as shown in Figure \(7.13 .\) The large cart is now forced to oscillate such that \(X=A \cos \omega t,\) with both \(A\) and \(\omega\) fixed. Set up the Lagrangian for the motion of the small cart and show that the Lagrange equation has the form $$\ddot{x}+\omega_{0}^{2} x=B \cos \omega t$$ where \(\omega_{\mathrm{o}}\) is the natural frequency \(\omega_{\mathrm{o}}=\sqrt{k / m}\) and \(B\) is a constant. This is the form assumed in Section 5.5, Equation (5.57), for driven oscillations (except that we are here ignoring damping). Thus the system described here would be one way to realize the motion discussed there. (We could fill the large cart with molasses to provide some damping.)

A mass \(m\) is suspended from a massless string, the other end of which is wrapped several times around a horizontal cylinder of radius \(R\) and moment of inertia \(I\), which is free to rotate about a fixed horizontal axle. Using a suitable coordinate, set up the Lagrangian and the Lagrange equation of motion, and find the acceleration of the mass \(m\). [The kinetic energy of the rotating cylinder is \(\frac{1}{2} I \omega^{2} .\)]

The "spherical pendulum" is just a simple pendulum that is free to move in any sideways direction. (By contrast a "simple pendulum"- unqualified - is confined to a single vertical plane.) The bob of a spherical pendulum moves on a sphere, centered on the point of support with radius \(r=R\) the length of the pendulum. A convenient choice of coordinates is spherical polars, \(r, \theta, \phi,\) with the origin at the point of support and the polar axis pointing straight down. The two variables \(\theta\) and \(\phi\) make a good choice of generalized coordinates. (a) Find the Lagrangian and the two Lagrange equations. (b) Explain what the \(\phi\) equation tells us about the \(z\) component of angular momentum \(\ell_{z^{*}}\) (c) For the special case that \(\phi=\) const, describe what the \(\theta\) equation tells us. (d) Use the \(\phi\) equation to replace \(\dot{\phi}\) by \(\ell_{z}\) in the \(\theta\) equation and discuss the existence of an angle \(\theta_{\mathrm{o}}\) at which \(\theta\) can remain constant. Why is this motion called a conical pendulum? (e) Show that if \(\theta=\theta_{0}+\epsilon,\) with \(\epsilon\) small, then \(\theta\) oscillates about \(\theta_{\mathrm{o}}\) in harmonic motion. Describe the motion of the pendulum's bob.

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